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Unformatted text preview: g ( x ) on all of R , we use that it is periodic. Thus, for any x,y R so that  xy  < (assuming without loss that < 1) we can substract 2 n for some n Z so that x2 n and y2 n are in [2 , 2 ]. Then,  sin xsin y  =  sin( x2 n )sin( y2 n )  < since  ( x2 n )( y2 n )  < . To show nonuniform continuity of fg on R , we shall use that sin x/x 1 as x and Theorem 5.4.2. Indeed, let x n = 2 n and u n = 2 n + 1 n . Then x nu n 0. x n sin x n = 0 0. But by the sine addition formula, u n sin u n = (2 n + 1 /n ) sin 1 /n . But 2 n sin 1 /n 2 . Thus for n suciently large,  fg ( x n )fg ( u n )  = (2 n + 1 /n ) sin 1 /n 1 / 2 . Hence by the Nonuniform Continuity Criterion, fg is not uniformly continous. 1...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 MetCalfe
 Calculus

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